Question

A certain metal M forms a soluble nitrate salt M NO Suppose the left half cell of a galvanic cell apparatus is filled with a 3.00 mM solution of M(NO) and the right half cell with a 3.00 M solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 20.0 °C.

1. Which electrode will be positive? What voltage will the voltmeter show?
a. left
b. right
2. Assume its positive lead is connected to the positive electrode. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.

Answer 1

Step-by-step explanation:

From the information given:

left half-cell = 3.00 mM M(NO)

right half-cell = 3.00 M M(NO)

Since the electrode are the same in both cells, then the concentration for the cell are also the same.

Negative electrode = Anode = lower concentration = 3.00 mM

Positive electrode = cathode = higher concentration = 3.00 M

Thus, right half cell will be postive electrode.

To determine the concentration cell:

`Ecell =\Big( (2.303* R* T)/(nF) \Big)log\Big(([cathode])/([anode])\Big)`

SInce [Cathode] > [anode],

`R = 8.314 J/K/mol, \\ \\ T = 273+20 = 293 K \\ \\ Faraday's constant (F)= 96500 C/mol`

n = 3

`Ecell ={ (2.303* 8.314* 293 )/((3* 96500))} log(3)/(0.003)`

`\mathbf{E_(cell) = 0.0582 V } \\ \\ \mathbf{E_(cell) = 0.06 V}`